The h Bar

@MAFIA36790 what do you mean?

user116211

@DHMO You called axiom.

@MAFIA36790 or rather a definition, since axiom is mathematical jargon

user116211

Ohm's Law is $\mathbf J = \sigma \mathbf E\,.$

bolbteppa

2:11 PM

Hmm, isn't Ohm's law an assumption in macroscopic electrodynamics for constant currents, arising Maxwell's equations imply you need a link between $\nabla \mathbf{j} = 0$ and $\nabla \times \mathbf{E} = 0$, Ohm's law is the assumption of a linear relation no?

What, isn't resistance defined to be the quotient when the voltage is divided by the current?

user116211

The general equation is $ J_x = \sum \sigma_{xj}~E_j $

wtf is this

PhD quantum electrodynamics

user116211

._.

user116211

You must have drunk Slereah's evil apple wine.

2:18 PM

I don't quite understand the next set of questions in the trigonometry book. It says "sketch the oriented arc on the Unit Circle which corresponds to the given real number." I know what the Unit Circle is, but I don't quite understand what it wants. The first problem is with $t = \frac{5\pi}{6}$.

Any help would be appreciated.

@heather I think they want you to draw the part of the circle that is that angle in radians (e.g. $2\pi$ would be the whole circle, $\pi$ half the circle)

Oh, so just draw the angle that number represents?

That's pretty easy - this would be almost half the circle. So a II quadrant angle.

Okay, thanks!

@ACuriousMind, how would you do that with $t=6$?

Well, for any $t$, $\frac{t}{2\pi}$ tells you how much of the full circle you need to draw

Internet work again!

I am a PhD in IT

user116211

hmm.

2:26 PM

@ACuriousMind Since you're doing math now, the hangover must be over

@heather has it been taught to you that "radian" is defined to be the ratio between, you know, the radius and the arc?

so you can do calculus with me pls :)

yesterday, by ACuriousMind

I don't even know what calculus is

@ACuriousMind, oh, okay, so it would be $\frac{6}{2\pi}$...oh, duh. Okay, thanks!

But go ahead and state your problem

2:27 PM

@heather precisely

$\Huge\color{purple}{\LaTeX}$

4

@ACuriousMind $E,F$ Banach, $z,x\in U\subset E$ open.

$f:U\to F$ differentiable on $U$

Let $[z,x]$ denote the line segment between $z$ and $x$. $U$ is convex so $[z,x]\subset U$.

I want to compute $$\lim_{x\to z}\left(\sup_{y\in[z,x]}||df(y)-df(z)||\right)$$

bolbteppa

@0celo7 go buck wild math.stackexchange.com/q/1971076/82615

operator norm

[Division by zero] Proposal # 5:

->test needed

@Secret we meet again

2:30 PM

indeed

@ACuriousMind I'm fairly certain for that limit to be 0, we need $f$ to be $C^1$, not just differentiable.

@Secret so q0=1 but 0q=0?

@ACuriousMind $df$ is the Frechet derivative.

@DHMO Actually, for this particular choice, it is impossible to prove 0*anything=0 without other mutlpicative inverses

@Secret sorry i did not understand this

2:32 PM

0q can =0 if you choose to knock out LDist and Lid+ instead

We are knocking out axioms that the zero anihilation theorems depend on to forbid zero from absorbing itself

This is why mapping the entire pathways of this proof is important

I see

well, if 0 is invertible then I don't see how there is any zero-annihilation theorem

@0celo7 Yes, I'd think so

Balarka Sen

Differential geometry is actually nice, even for surfaces in R^3.

exactly, if zero is at least one sided invertible, then zero cannot absorb things, in paritcular: zero cannot absorb itself

The key to divide by zero is to make $0^2\neq 0$

and this is why division by zero algebra break so many rules except for relatively trivial cases

@Secret why is this necessary?

2:35 PM

That will be illustrated on the next slide, but short answer is imagine what will happen if you operate q both sides on that equation

Currently we identified the following common places where proposed division by zero algebra will collapse:

In particular, if any powers of zero is equal to zero, one can induce collapse into the trivial ring by operating q

@Secret you know, you just need to remove the associativity

that is not enough, because $0^2=0$ does not rely on associativity

as long that equation is true, your algebraic system will collapse

Can someone tell me if I did the following problems right:

However, we do found except for a class of trivial cases, associativity is almost always broken

2:40 PM

The following system is the only class of cases where zero have a legitimate multiplicative inverse:

@heather q8: reference angle should be positive, right

@heather other questions are good

@DHMO, oh, duh...so would it be $\frac{\pi}{12}$ then?

This is ensured because the + cayley table act as a forgetful functor (according to tobia) thus all associative and distributive laws become trivially satisfied

@ACuriousMind Thanks.

@heather precisely

2:42 PM

and that bypass the collapse problem, resulting in all the left zeros to have mutlpicative inverses

@DHMO, I can't think today. =) Thanks!

@heather you are welcome

Victory! Week 1 is complete!

typo: only class of cases we knew

=)

2:44 PM

These class of counterexamples hence proved that division by zero is actually possible for nontrivial algebraic structures

@Secret commutative addition: 0+0 = 1+q = q+1 = 0; 0+q = 1+1 = q+1 = q; q+1 = 1+0 = q+q = 1

@ACuriousMind this man looks CLEAN

@DHMO how do you get 1+1=q from the above rules?

@Secret no, i defined another addition

o then, its a different algebraic system

2:47 PM

yes

because i'm not satisfied with non-commutative addition

Well, I am not sure if division by zero algebra can handle commutative addition. There are so many ways to choose which axioms to break in order to block off $0^2=0$. The LDist, Lid* I have not tested it before, thus it might be possible to restore commutativity there

sorry, typo now fixed

I am not sure if mathematicians actually do that, cause what I knew them better is proving things and finding counterexamples, but not purposely breaking axioms to make new structures (Well they do relax axioms, but having this as a primary goal I am not sure if there are examples)

Q: Removing electron from inner shell of an atom

-1

Can we remove an electron from the inner shell of an atom without disturbing the outer orbital? If so, how?

electrons atomic-physics orbitals

Thoughts on this question?

...I do have an engineering mindset when it came to mathematics, by trying to build something by taking things apart

@Secret why do we need to block off axioms?

2

because some theorems when proven true will exclude other things for a given mathematical structure

e.g. the moment you have zero absorbing itself, only the trivial ring can be a division by zero algebra

FAQ: Wheels are not good enough, they don't actually have a mutiplicative inverse of zero as they have shown that attempt to do so will immediately collapse it into the trivial wheel

and the pseudoinverse in meadows is not actually even a pseudoinverse because it can be shown there $0^{\sim 1}=0$

I still yet to understood Jonathen Cender's near zero approach, though

They are the only 3 known serious attempt at diving by zero, with wheels being the most sucessful of them because they actually have "division by zero" terms

but none of them has a legitimate inverse of zero, thus technically speaking they have not divided by zero

2:59 PM

@Secret just name it infinity and make it an absorber

well then you end up with what is bascially the IEE floating point system, where NaN absorbs everything

(having said that I have not seriously checked what happens if the inverse of zero becomes an absorber. I think it might be possibel to prove that $1=\infty$ there)

and that is a collapse

Proposal # 4 look like this (with the associative problem remains to be solved)

@Secret nope, not interesting

@Secret I know how to make zero invertible: rename it as one and rename one as zero

@DHMO nope, that is just relabelling: A zero element is one that either act as addiative identity or absorbs (at least one sided)

@Secret your zero doesn't look like an additive identity to me

@ACuriousMind As much as it pains me to say this...the game is very good.

Q: It is possible to show/prove that the cancellation property is necessary to prove $0x=0$ for $x\not\in \mathbb{Z}^+ \cup \{0\}$?

3:09 PM

because in that system, it is only a right additive identity, that is $a+0 = a$, but $0+a\neq a$

However because you prefer commutative algebra, proposal # 4 will not look appealing to you. Not to mention it has a serious nonassociative problem that an axiom is needed to limit the growth of the nonassociative terms

(all of this is based on discussion with tobias and I analysing that system)

Let's see if proposal # 5 can restore commutativity (which is the current one on my table right now)

Side note: the squared rock paper scissor commutative magma is accidentally (re?)discovered when building proposal # 4 and make it finite size to easily test for collapse

other things (which might be well known) is that one sided elements tend to break associativity

In this MSE, we found that additive identities are nonunique unless addition is associative. We also found that cancellation is unnecessary for proving the zero anihilation theorems

3

For any nontrivial algebraic structures with additive identity 0 and multiplicative identity 1 (and binary operation defined by "juxaposition of its arguments"), and at least one sided distributive law holds, one can easily prove that 0 is an absorber, i.e. $0x=0$ for $x=\{0,1\}$ as follows: $$0...

abstract-algebra proof-writing

Damn... I always think I don't understand a damn about quantum mechanics, and I am always right

I can't believe I got that wrong all the way till I read this

edge.org/response-detail/25453

Well, I handle the transition and tunneling cases as the amplitudes of the eigenfuctions that made up the wavefunction changes with time, and that's how an electorn beocme more or less liekyl to be foudn smoewhere

@Shing, hey, but you figured it out, right?

So in a sense, one cannot really say an electron have tunneled through a barrier, because its wavefunction has already spread to the otherside, and we just found it on the other side when measuring it

user218912

I should just play psp games on my ps vita because all the new games are so bad...

3:22 PM

@heather yeah, the good news is I know it now.

@Secret frankly, I still have a hard time understanding uncertainty principle... I can do the math, but I just don't understand it.

@Shing, maybe I could attempt to explain it, because I think I sort of get it.

"no physics intuition", I think that's what they say

@heather sure, I would love to hear it from you

@Shing, okay. So if you have a nice particle, we'll just say an electron.

@0celo7 hehe

The key idea is that, "the order matters". If you measure one observable A and then B, the state is projected to an eigenstate of A, and then when you measure it again, it projected into some eigenstate of B. The fact that trying to gain information about one of the observable will project the state down to an eigenstate (hence changing it) is the reason of the uncertainty

the order only does not matter if all the eigenstate in question is for both observables A and B

3:25 PM

we know that the more accurately we measure position the less accurately we can measure momentum, and vice versa.

user218912

@heather okay but how do we know that? xD

@bloo, geesh, I'm getting to that. =)

user218912

you can't understand the uncertainty principle, it's just something that occurs as a consequence of the math that describes qm.

user218912

it's not intuitive, so if you understand the math part you're good.

@heather Obligatory statement: The uncertainty principle has nothing to do with the "accuracy" of measurements.

Swapnil Das

3:28 PM

@bl00 Why aren't most stuff in QM, intuitive?

user218912

because we evolved not to understand quantum things since we're so much larger than them.

@ACuriousMind, yes, that was poor phrasing, sorry.

thanks all for your help and kind words, I guess my only path to salvation is doing more problem-sets....

@bl00 Ah, the next common misphrasing: How relevant quantum mechanics is to a phenomenon has not intrinsically to do with its size.

user218912

@ACuriousMind what?

3:31 PM

e.g. Our sun shines because of tunnelling. Also plants use entanglement to make photosynthesis extremely efficient

user218912

oh I see.

user218912

yeah but they're still quantum processes.

It's better to say that in everyday situations, the world is well-approximated by classical thinking. But this is not purely due to the "size" of objects

user218912

there can be large scale quantum processes by they still originate at a quantum level.

also superconductor

3:33 PM

@Shing, does that make sense? The above explanation?

user218912

@heather he said it has nothing to do with measurement, using the word observe instead of measure doesn't change anything.

@heather You're explaining something different from what the uncertainty principle actually says. The uncertainty principle is not about the measurement affecting the system.

@ACuriousMind, oh, dear. I'll delete that then. But honestly, that's what my physics textbook said.

Although every measurement affects the system in quantum mechanics, this is not what the uncertainty principle is about

@ACuriousMind is it really not?

user218912

3:34 PM

@heather you don't need to delete it. ._.

um... that makes sense to me. but I can't rephrase them in my own words, so I guess I still don't understand. never mind, I had not understand degeneracy in quantum physics untill I worked some touch problem set about it.

@ACuriousMind, then what is it, then?

ACM: ok, in that case, it seems my explanation is also wrong then, because I use measurement

user218912

@heather what you said is right but it is not the general uncertainty principle, it is a high school level description of it applied to measurement.

@ACuriousMind Vault 11 tho

what the hell happened here

3:36 PM

@Sanya No. The uncertainty principle is about the fact that if I take a state and do position and momentum measurements on many identical copies of it, then the narrower the range of positions I get is, the broader the range of momenta is, and vice versa

Okay, here's the other thing my textbook says, tell me if this is right. It says that quantum uncertainties stem from the wave nature of matter, and that a wave occupies some space and some time by its nature. So we can't squeeze that wave to a point in space or a single instant of time because then it wouldn't be a wave. So this fuzziness leads to a fuzziness in measurement.

@ACuriousMind that's the clearest interpretation I have ever read

(we do position measurements on one half of the copies and momentum measurements on the other half, we are not doing the measurements in succession)

@ACuriousMind sacrificial chamber?!??

Wtf is going on

user218912

@heather that doesn't mean anything though, where's the math to back it up?

3:39 PM

@0celo7, we're talking about the uncertainty principle and you're talking about sacrifical chambers. Ironic juxtaposition =P

AHHHHHHHH

ACM HELP WHAT IS HAPPENING

@0celo7 Let's just say VaultTec wasn't nice to all the vaults

@bl00, that's just what the book says. I guess I thought it meant that since an electron is represented by a wave it cannot be confined to one point in space or one instant in time?

And so we can't measure it as such

lol I got destroyed

@ACuriousMind ok but why was there a sacrificial chamber

Insert pure confusion about the uncertainty principle here.

3:42 PM

@heather it's just Cauchy-Schwarz

@ACuriousMind QM is always disturbing

@heather Ehhhh, that's not exactly wrong (there are indeed no states with definite position or momentum), but I don't like saying that's due to the "wave nature of matter". I could show that that is the case in quantum mechanics without ever mentioning the word "wave".

@0celo7, uh...what are you talking about?

@0celo7 Did you read the stuff in the terminals? It should become clear from that

user218912

@heather the uncertainty principle comes from the cauchy-schwarz identity, if you have 2 noncommuting observables.

3:43 PM

@ACuriousMind, my textbook has then only increased my confusion.

Yes, all non-technical accounts of quantum mechanics tend to use outdated concepts and only increase confusion, in my experience

@Sanya I don't find it particularly disturbing

ACM: Is there a name for what I mentioned about measurement in succession being dependent on the order it is done?

@ACuriousMind, okay. I read your explanation, and I'm not sure I understand. Why is there that spread? Is there no explanation, it just is?

user218912

@heather no there is an explanation.

@Secret It's a consequence of the measurement affecting the state, I don't think it has its own name

3:46 PM

@ACuriousMind well, it's again "spooky action at a distance", a bit like the Bell experiment. I mean, let's say we have 100 identical copies, we split them, do measurements on the first half - suddenly for the second half the scatter of the measurement is predetermined

@bl00, what is it?

ok

user218912

@heather it should be in any advanced book on quantum mechanics.

@heather It's an axiom of quantum mechanics that the results of measurements are not always the same for the same state, but are distributed according to a probability density defined by the state.

There is no explanation for why measurements are probabilistic like that within QM.

@ACuriousMind, okay, there isn't an explanation.

3:48 PM

@ACuriousMind not all of them, no

Boone can't die, right?

@bl00 Hmmm?

he's not showing up any more

I might have told him to wait somewhere

@bl00, I don't exactly have an advanced book on quantum mechanics handy...

what do?

@ACuriousMind, @Secret, would it be called the "observer effect" (Wikipedia mentions that)?

It "refers to changes that the act of observation will make on a phenomenon being observed"

3:50 PM

@Sanya What do you mean, the "scatter is predetermined"? What's determined is the standard deviation of an ideal, infinitely often repeated measurement (even before you make the first measurement). This does not determine any particular measurement result.

@0celo7 Well, I think you can find out that Vault 11 was an experiment to see whether people would kill one of their own each year if told that the vault would otherwise kill them all.

user218912

@ACuriousMind what?

@bl00 You said there is an explanation. I don't know what you are referring to.

@ACuriousMind wait, we just said, to quote you "if I do position measurements with a certain uncertainty/standard deviation on some of the copies, this influences the standard deviation with which I can measure momentum on the other copies", right?